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Question Number 185953 by ajfour last updated on 30/Jan/23

Commented by ajfour last updated on 30/Jan/23

$$Woulditber\approx 0.432?$$

Commented by TUN last updated on 30/Jan/23

$$S\left(triangle\right)=p\times r$$$$<=>\frac{1}{2}\sqrt{1+r^2}\times \sqrt{\frac{1}{x^2}+1}=\frac{\sqrt{1+r^2}+\sqrt{\frac{1}{r^2}+1}+r+\frac{1}{r}}{2}\times r$$$$<=>r=0.4320408003$$$$$$

Commented by ajfour last updated on 30/Jan/23

$$thanks,Iunderstand.wow!$$

Answered by mr W last updated on 30/Jan/23

Commented by mr W last updated on 30/Jan/23

$$\tan \theta =\frac{r}{1}=r$$$$\cos \theta =\frac{r+\frac{r}{\tan \frac{\theta }{2}}}{\frac{r}{\tan \frac{\theta }{2}}+\sqrt{1+r^2}-r}$$$$\cos \theta =\frac{1+\frac{1}{\tan \frac{\theta }{2}}}{\frac{1}{\tan \frac{\theta }{2}}+\sqrt{1+\frac{1}{{\tan }^2\theta }}-1}$$$$lett=\tan \frac{\theta }{2}$$$$\frac{1-t^2}{1+t^2}=\frac{1+t}{1-t+t\sqrt{1+{\left(\frac{1-t^2}{2t}\right)}^2}}$$$$t^3-t^2+5t-1=0$$$${(s+\frac{1}{3})}^3-{(s+\frac{1}{3})}^2+5(s+\frac{1}{3})-1=0$$$$s^3+\frac{14s}{3}+\frac{16}{27}=0$$$$s=\sqrt[3]{\frac{2\sqrt{78}}{9}-\frac{8}{27}}-\sqrt[3]{\frac{2\sqrt{78}}{9}+\frac{8}{27}}$$$$t=\frac{1}{3}(1+\sqrt[3]{6\sqrt{78}-8}-\sqrt[3]{6\sqrt{78}+8})$$$$\approx 0.206783$$$$\Rightarrow r=\frac{2t}{1-t^2}\approx 0.432$$

Commented by ajfour last updated on 30/Jan/23

$$ThankyouSir.Allcorrect!$$

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